leetcode - Single Number

leetcode - Single Number

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int res = 0;
        for(int i = 0; i < nums.size(); i++){
            res ^= nums[i];
        }
        return res;
    }
};

普通的解法无需再提，这里需要线性复杂度，不使用额外内存的解法：

利用异或运算的性质：

a^b = b^a 

a^a = 0

0^a = a

显然，将所有数字异或之后，出现两次的数字都变成了0，只剩下出现一次的数字。

leetcode - Single Number

原文：http://www.cnblogs.com/shnj/p/4739392.html